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\begin{document}

\section{Problem Sheet 1}

\subsection{Capacitor}

\subsubsection{a)}

Let the capacitor have area $A$, with plate separation $d$. The volume of the inside of the capacitor is thus $V = Ad$.

Assume that $\E$ field is zero outside the capacitor.

The momentum density stored within the fields is $$\vec p(\r) = \frac{1}{c} \E(\r) \times \B(\r)$$ hence the total momentum is the density integrated over the region where $\vec E \times \vec B$ is non-zero (i.e. the inside of the capacitor) is simply $$\vec P = \int \d V' \vec p(\r) = \frac{Ad}{c} \E \times \B = \frac{Ad}{c} EB \vec e_z \times \vec e_x = \frac{Ad}{c} EB \vec e_y$$

The force on a conductor $l$ is $$\vec F = \frac{1}{c} \int_l \d r' \J(\r') \times \B(\r')$$

The current $\J(\r)$ at any point in the conductor is $$\J = \diff{Q}{t} \vec e_z$$ where $Q$ is the charge on the plates of the capacitor. Hence the force reduces to $$\vec F = \frac{1}{c} \int_l \d r' \diff{Q}{t} \vec e_z \times B \vec e_x = \diff{Q}{t} \frac{Bd}{c} \vec e_y$$

The total impulse is the force integrated over the time taken for the capacitor to discharge $$\vec P = \int \d t \diff{Q}{t} \frac{Bd}{c} \vec e_y = \diff{}{t}\int \d t \frac{Q B d}{c} \vec e_y = \frac{QBd}{c} \vec e_y$$

Now compare with the previous expression for $\vec P$. Insert the definition of capacitance $Q = CV$ into the above expression, remembering that for a parallel plate capacitor $C = A / d$, $V = E d$, to obtain $C = A E$. Thus $$\vec P = \frac{AdEB}{c} \vec e_y$$ as required.

\subsubsection{b)}

The wire is now absent, hence the force must be exerted directly upon the charged plates themselves. The changing magnetic field will induce a 	

\section{Boundary Impedance}

\section{Monopoles}

Angular momentum density is

$$j_{em}(\r) = \r \times \vec p(\r) = \frac{1}{c} \r \times \left( \E(\r) \times \B(\r) \right)$$

hence the total angular momentum (measured about the origin) is this quantity, integrated over the extent of the fields (i.e. all space)

$$J_{em} = \frac{1}{c}  \int \d V'  \r' \times \left( \E(\r') \times \B(\r') \right)$$

Now insert explicit expressions for $\E = q (\r - \vec a) / (4 \pi | \r - \vec a |^3)$ and $\B(\r) = g \r / (4 \pi r^3)$

$$J_{em} = \frac{qg}{16 \pi^2 c}  \int \d V'  \r' \times \left( \frac{ (\r' - \vec a) \times \r' }{ |\r' - \vec a|^3 r^3 } \right)$$

\section{Problem Sheet 2}

\subsection{Gauge Transformations}

$\E = \grad \phi$, $\phi = 0$ implies that $\E = \grad 0 = 0$.

\begin{eqnarray*}
\B &=& \curl \vec A = -\frac{q}{4 \pi} ct \curl \frac{ \r }{ r } \\
&=& -\frac{q}{4 \pi} ct \left( \frac{1}{r} \curl \r + \r \times \grad (1/r) \right) \\
&=& -\frac{q}{4 \pi} ct \frac{1}{r^2} \r \times \r = 0
\end{eqnarray*}

(ii)

Gauge transformations:

$$\vec A \rightarrow \vec A + \grad \chi$$

$$\phi \rightarrow \phi + \frac{1}{c} \pdiff{\chi}{t}$$

So for given $\phi, \vec A, \chi$:

$$\grad \chi = \frac{q}{4 \pi} ct \frac{1}{r^3} \r$$

$$\vec A' = \frac{q}{4 \pi} ct \left( \frac{1}{r^3} \r - \frac{1}{r^3} \r \right) = 0$$

$$\phi' = - \frac{q}{4 \pi} \frac{1}{r}$$

The magnetic field $\B$ is clearly still zero. The electric field $E$ can be checked 

(iii)

Lorentz gauge requires that 

$$\div \vec A + \frac{1}{c} \pdiff{}{t} \phi = 0$$

and $\vec A$ and $\phi$ are \emph{already} in Lorentz gauge. Substituting the transformed potentials $\vec A' = \vec A + \grad \chi$ and $\phi' = \phi + \frac{1}{c} \pdiff{\chi}{t}$ into the Lorentz condition gives

$$\div \left(\vec A + \grad \chi \right) + \frac{1}{c} \pdiff{}{t} \left( \phi + \frac{1}{c} \pdiff{\chi}{t} \right) = 0$$
$$\div \vec A + \div ( \grad \chi ) + \frac{1}{c} \pdiff{\chi}{t} + \frac{1}{c^2} \pdiffsq{\chi}{t} = 0$$

and since the terms involving the original potentials $\vec A$ and $\phi$ sum to zero, we are left with the condition

$$\laplacian \chi + \frac{1}{c^2} \pdiffsq{\chi}{t} = 0$$

(since $\div ( \grad f ) = \laplacian f$ )i.e. a wave equation for $\chi$. Hence any periodic $\chi$ satisfying the wave equation will do. 

\subsection{Coulomb Gauge}

Maxwell's equations (in Lorentz-Heaviside):

$$\div \B = 0$$
$$\curl \B = \frac{1}{c} \pdiff{\E}{t} + \frac{1}{c} \J$$
$$\div \E = \rho$$
$$\curl \E = - \frac{1}{c} \pdiff{\B}{t}$$

Rewrite the curl equation for $\E$

$$\curl \left( \E +  \frac{1}{c} \pdiff{}{t} \A \right) = 0$$

and since the curl of the terms in brackets vanishes, this implies that they can be rewritten as the gradient of a scalar $\phi$

$$\E = - \grad \phi -  \frac{1}{c} \pdiff{}{t} \A$$

Now, using existence of vector potential $\B = \curl \A$, and the above expression for $\E$, rewrite the inhomogeneous magnetic equation

$$\curl ( \curl \A ) = \frac{1}{c} \pdiff{}{t} \left( - \grad \phi -  \frac{1}{c} \pdiff{}{t} \A \right) + \frac{1}{c} \J$$

Apply the expansion for the left-hand side, and write $\J = \grad \dot \phi + \J_2$

$$\grad ( \cancel{ \div \A } ) - \delsq \A = - \frac{1}{c^2} \pdiffsq{\A}{t} + \frac{1}{c} \J_2 $$

$$ \delsq \A - \frac{1}{c^2} \pdiffsq{\A}{t} = - \frac{1}{c} \J_2 $$

where $\div \A = 0$ is the Coulomb condition. This is an inhomogeneous wave equation for $\A$, which depends solely on $J_2$.

Using the inhomogeneous equation for $\E$

$$ \div \left( - \grad \phi -  \frac{1}{c} \pdiff{}{t} \A \right) = - \rho $$

$$\Rightarrow \delsq \phi + \frac{1}{c} \pdiff{}{t} \cancel{ \div \A } = \rho$$

$$\Rightarrow \delsq \phi = \rho$$

which is Poisson's equation for $\phi$

Since $\J_1 = \grad \dot \phi$ is the gradient of a scalar, its curl must vanish

$$\curl \J_1 = \curl \grad \dot \phi = 0$$

Proof:

Let $\vec f = \grad \psi$

$$\left[ \curl \vec f \right]_i = \epsilon_{ijk} \partial_j \partial_k \psi = - \epsilon_{ikj} \partial_j \partial_k \psi $$

where we have permuted the indices of the antisymmetric tensor. But partial derivatives commute, i.e. $\partial_j \partial_k \equiv \partial_k \partial_j$ hence we can switch the order of derivatives, and then relabel the dummy indices $k \leftrightarrow j$ to obtain

$$ \epsilon_{ijk} \partial_j \partial_k \psi = - \epsilon_{ijk} \partial_j \partial_k \psi $$

and so $\vec f$ must vanish.

Taking the divergence of the wave equation for $\A$

$$\div \delsq \A - \frac{1}{c^2} \pdiffsq{}{t} \cancel{ \div \A } = - \frac{1}{c} \div \J_2$$

the Laplacian of $\A$ consists of two terms $\curl (\curl \A) - \grad ( \div \A )$. The divergence term is zero by the Coulomb condition, and the divergence of a curl always vanishes, to give

$$\div \J_2 = 0$$

Proof:

Let $\vec f = \curl \vec g$

$$\div \vec f  =  \partial_i \epsilon_{ijk} \partial_j g_k =  - \partial_i \epsilon_{jik} \partial_j g_k $$

and similarly to above, by switching order of differentiation and relabelling dummy indices $i, j$

$$ \partial_i \epsilon_{ijk} \partial_j g_k =  - \partial_i \epsilon_{ijk} \partial_j g_k $$

and so $\vec f$ must vanish.

Using the Poisson equation for $\phi$

$$\delsq \phi = \rho$$

Realise that current conservation 

2.4

A mass falling under gravity has a trajectory given by

$$\vec z(t) = \hat{\vec z} \left( z_0 - \frac{1}{2} g t^2 \right)$$

Taking the origin to be immediately below the falling particle, then considering the charge and its image to be a dipole, the dipole moment of the configuration is

$$\vec p = q \vec z(t)$$

Larmor's formula gives the total power radiated by a dipole

$$P = \frac{\ddot p^2}{6 \pi c^3}$$

inserting the relevant dipole moment $\ddot{ \vec p} = - qgt \hat{\vec z}$

$$P = \frac{q^2 g^2}{6 \pi c^3}$$

2.5

Taking the origin at the equilibrium point of the particle, the trajectory of the oscillating particle is

$$\vec z(t) = - \hat{\vec z} d \cos( \omega t )$$

where $\omega = \sqrt{ k / m }$. As before, we consider the particle as a dipole with moment

$$\vec p = q \vec z(t)$$

The Poynting vector for dipole radiation is

$$\vec S = \frac{1}{16 \pi^2 c^3 r^2} \ddot{p}^2 \sin^2\theta \hat{\vec r}$$

again, inserting the relevant dipole moment $\ddot{\vec p} = \hat{\vec z} d \omega^2 \cos (\omega t)$, the power radiated is

$$\vec S = \frac{1}{16 \pi^2 c^3 r^2} d \omega^4 \cos^2 (\omega t)  \sin^2\theta \hat{\vec r}$$

a)

In our chosen geometry, $r^2 = R^2 + h^2$ (i.e. a right triangle), and so $r = h \cos \theta$, and the radiation impinging on the floor at angle $\theta$ is

$$\vec S = \frac{1}{16 \pi^2 c^3 h^2} d \omega^4 \cos^2 (\omega t)  \tan^2\theta \hat{\vec r}$$

$\tan^2 \theta$ is maximised for $\theta = \pi / 2$, hence the intensity of radiation is maximised where $\theta = \pi / 2 \Rightarrow R = h$. This is as expected, recalling the toroidal shape of the surface of equal energy density around an oscillating dipole.

b)

For the total radiation impinging on a floor of infinite extent, we have the Larmor formula, with an additional factor of $1/2$ (since we are integrating over a half-sphere, rather than a full sphere)

$$\langle P \rangle = \frac{w^4 p_0^2}{12 \pi c^3}$$

where $w = \sqrt{k/m}$ is the angular frequency of the oscillation, and $p_0 = d$ is the amplitude of the oscillation, hence in our case

$$\langle P \rangle = \frac{k^2 d^2}{12 \pi m^2 c^3}$$

c)

A mass undergoing SHM with maximum amplitude $d$ has total energy

$$E_0 = \frac{1}{2} m \omega^2 d^2$$

and with energy lost at a rate $\langle P \rangle$, after a time $\tau$

$$E_0 \left(1 - \frac{d}{e} \right) = \tau \langle P \rangle$$

Using the Larmor formula from (b) for $\langle P \rangle$, and solving for $\tau$

$$\tau = $$

